how to solve the prob/

Question

The average size (in square feet) of homes built in the United States in the fourth quarter of 2008 was 2,343. Assume the variable is normally distributed and the standard deviation is 152 square feet. Find the percentage of homes that will have between 1,900 and 2,500 square feet. Give your answer as a percent rounded to two decimal places (to the nearest hundredth)

Answer 1

Once again a standard problem that can be solved by looking up a good introductory text.  But I shall help you out time.

 

Let Y = Area 9in sq. ft.) of homes built in the United States in the fourth quarter of 2008.

 

We are given that Y is normally distributed with mean mu = 2343 and sigma = 152.

 

We want to find the proportion (percentage) P of homes the areas of which are between 1900 and 2500.  Symbolically, we want to find P [ 1900 < Y < 2500].  Since the normal is a continuous distribution, we can ignore or include the equality signs with no problem at all.  (If you are confused, ask me as a comment.)

 

The following steps are an exercise in inequalities.  The idea is to convert the above proportion in terms of "z_values" of the normal curve.

 

Subtracting mu = 2343 in the string of inequalities above we get

 

1900 < Y < 2500 is equivalent to 1900 - mu < Y - mu < 2500 - mu, i.e., 1900 - 2343 < Y - 2343 < 2500 - 2343, i.e,  -443 < Y  - 2343 <  157.

 

Now, dividing the last string of inequalities by the std dev sigma = 152, we get (since we are dividing by a positive number, the inequality signs do not change), we get

 

-443/152 < (Y - 2343)/152 < 157/152, i.e., -2.91447 < (Y - 2343)/152 < 1.03829.

 

The quantity in the middle is the variable Z = ( Y  - mu) / sigma which has a standard normal distribution with mean 0 and sigma = 1.  So to solve the problem, we have to find the percentage of area under the standard normal curve between -2.91447 and 1.03829.

 

These areas are well-tabulated and, are, in fact available on the internet in the public domain.

 

For example, you can use the web-link http://www.danielsoper.com/statcalc3/calc.aspx?id=2

 

When you go on to the website you will see that you get the area from minus infinity to the "z-score" you specify  (this is the are to the left of the z-score you specify.

 

So, to arrive at the answer you want, you find (a) the area to the left of 1.03829, and SUBTRACT from it the area to the left of (b) -2.91447.  (Heather, I would like you to draw a picture here and understand this step fully.  It is no hard at all.)

 

From the web-link I get, (a) area to the left of  1.03829 = 0.8504,

                                   (b) area to the left of -2.91447 = 0.0018  ( I had to input -2.9145 as z-score),

to 4 decimal places.

 

he difference as stated above is 0.8486.  Therefore, the answer to your problem, IN THE CONTEXT of the problem is:

 

The percentage of homes built in the United States in the fourth quarter of 2008, that have areas between 1900 and 2500 sq. ft., is 84.86, to two decimal places.

 

Heather, I would like you to compute the percentage of such homes with areas between 2191 and 2647 sq ft.  Post your answer as a comment.  Then and only then I will be happy for writing this answer for your sake!

 

Answer 2

(1900-2343)/152= -2.91.     from tables area = 0.4982

 

(2500-2343)/152= 1.0329.    from tables. area = .3485

 

Total area = 0.4982+ 0.3485 = 0.8467

 

In percentage terms = 84.67%

Answer 3

Heather the trick here is to convert values to z look up values so you can look up percentages on a z table. For example the z value at 1900 is the 1900 less the average all divided by thestandard deviation. In this case (1900-2343)/152. Likewise the z value at 2500 is (2500-2343)/152. You will find the z value ( and percentage) larger for 2500. Don'tsubtract the z values! Find the percentage of each z value and then find the difference between percentages. 

 

Hope that helps,

Deanna